Question

In $\Delta ABC,\angle A={45}^{\circ },\angle B={60}^{\circ }$ and $\angle C={75}^{\circ }$. The angles of the triangle formed by joining the mid-points of the sides of this triangles are

• A. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$
• B. ${45}^{\circ },{60}^{\circ },{75}^{\circ }$
• C. ${70}^{\circ },{70}^{\circ },{40}^{\circ }$
• D. ${40}^{\circ },{90}^{\circ },{40}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ },{60}^{\circ },{75}^{\circ }$


Given F is the midpoint of AB.
So $AF=\frac{1}{2}AB$

As E and D are midponits of AC and BC respectively.
$ED=\frac{1}{2}AB$ and ED||AB.

Similarly AE =DF and AE||DF.

So , AFDE is a parallelogram.

Similarly BFED and CDFE are parallelograms.

The measures of the angles of the $\Delta DEF$ formed by joining the mid-point points of the sides of the $\Delta ABC$ is

$\angle D=\angle A$
[Since AFDE is a parallelogram]
$\angle E=\angle B$
[Since BFED is a parallelogram]
and $\angle F=\angle C$
[Since CDFE is a parallelogram]
Hence, the angles of the triangle formed by joining the mid-points of the sides of this triangles are ${45}^{\circ },{60}^{\circ },{75}^{\circ }.$